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受欢迎的微积分 >

integral of (-2^{3-2x})

解答

\int (- 2^{3 - 2 x}) d x

解答

\frac{2 ^{2 - 2 x}}{ln ( 2 )} + C

求解步骤

\int - 2^{3 - 2 x} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int 2^{3 - 2 x} d x

2^{3 - 2 x} = 2^{3} \cdot 2^{- 2 x}

= - \int 2^{3} \cdot 2^{- 2 x} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 2^{3} \cdot \int 2^{- 2 x} d x

化简

= - 8 \cdot \int 2^{- 2 x} d x

使用换元积分法

= - 8 \cdot \int - 2^{u - 1} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 8 (- \int 2^{u - 1} d u)

化简

2^{u - 1} : 2^{u} \cdot 2^{- 1}

= - 8 (- \int 2^{u} \cdot 2^{- 1} d u)

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 8 (- 2^{- 1} \cdot \int 2^{u} d u)

使用指数法则:

a^{- 1} = \frac{1}{a}

= - 8 (- \frac{1}{2} \cdot \int 2^{u} d u)

\int a^{x} d x = \frac{a ^{x}}{ln a}

= - 8 (- \frac{1}{2} \cdot \frac{2 ^{u}}{ln ( 2 )})

u = - 2 x

代回

= - 8 (- \frac{1}{2} \cdot \frac{2 ^{- 2 x}}{ln ( 2 )})

化简

- 8 (- \frac{1}{2} \cdot \frac{2 ^{- 2 x}}{ln ( 2 )}) : \frac{2 ^{2 - 2 x}}{ln ( 2 )}

= \frac{2 ^{2 - 2 x}}{ln ( 2 )}

解答补常数

= \frac{2 ^{2 - 2 x}}{ln ( 2 )} + C

作图

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